ENGINEERED IN NAIROBI. WORKING ACROSS EAST AFRICA.

SOLUTIONS / APPLICATION GUIDE

Maintenance, repair & operations

Prepare shutdowns and replacement decisions with a clear asset record. Compare motor energy costs and bring fault evidence to the maintenance conversation.

Illustrative maintenance, repair & operations equipment
AI-generated equipment illustration.

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CHOOSE YOUR APPROACH

Start with the application.

Condition review

Record symptoms, alarms, load conditions and inspection history before deciding what to replace.

Planned shutdown

Define isolation boundaries, permits, spares, tests and the return-to-service sequence.

Repair or replace

Compare condition, compatibility, downtime, energy use and supportability.

BUILD A USEFUL BRIEF

Bring these details.

  • Asset register, nameplates and single-line diagrams
  • Fault history, alarms, measurements and photographs
  • Shutdown window and operational constraints
  • Required spares, test records and acceptance criteria
Send your project brief ↗
Can equipment be serviced while energised?

Maintenance should be planned around approved isolation and a competent risk assessment. A website estimate never authorises live work.

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COMPONENTS THAT COMPLETE THE APPLICATION

Specify the whole working system.

Start with the equipment duty. Add the control, protection and measurement functions that your project needs; these are options to review, not a pre-approved assembly.

Images are AI-generated illustrations of equipment categories, not LOVATO model photographs. Confirm dimensions and ratings from the selected model datasheet.

AI-generated illustration of the broader equipment category; not a manufacturer model

Motor protection & control

Contactors

Switch motors or other loads using the appropriate utilisation category.

Confirm: Load type and duty · coil supply · auxiliary contacts.

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METHODS & WORKED EXAMPLES

Understand the estimate.

These examples use the default inputs. Change the live controls above for your own scenario; the reference examples below remain fixed.

Motor energy comparison — worked example

Annual kWh = rated shaft kW × load fraction × annual hours / efficiency fraction. Cost difference uses the entered energy tariff.

Example inputs

  • Rated shaft power: 30 kW
  • Shaft load: 75 %
  • Existing efficiency at duty: 88 %
  • Replacement efficiency at duty: 93 %
  • Annual running hours: 4000 h
  • Energy tariff assumption: 25 KES/kWh
  • Installed replacement cost: 150000 KES

Example result

  • Existing annual energy: 102,272.7 kWh
  • Replacement annual energy: 96,774.2 kWh
  • Annual cost difference: 137,463 KES

Assumptions & limits

  • Both motors deliver the same shaft duty. Use efficiency at this load, not an unrelated nameplate point.
  • Tariff is your scenario input, not a published utility rate.
  • Simple payback excludes finance, maintenance, downtime and tariff changes. Negative difference means higher operating cost.
  • Preliminary planning only. Final selection needs project-specific engineering verification.

Technical reference: US Department of Energy: motor systems guide. This reference does not certify the website calculation or an NPE assembly.

CT current conversion — worked example

Ideal secondary current = primary current × rated secondary / rated primary. This does not evaluate CT burden, accuracy or saturation.

Example inputs

  • CT rated primary: 400 A
  • CT rated secondary: 5 A
  • Primary current: 240 A

Example result

  • CT ratio: 80:1
  • Ideal secondary current: 3 A
  • Primary rating utilisation: 60 %

Assumptions & limits

  • Isecondary = Iprimary × rated secondary / rated primary.
  • Ideal ratio only; no allowance for saturation, phase error, accuracy class or burden.
  • Never open-circuit an energised CT secondary. Protection settings require a coordination study and approved testing.
  • Preliminary planning only. Final selection needs project-specific engineering verification.

Technical reference: Schneider Electric: current transformer guidance. This reference does not certify the website calculation or an NPE assembly.

kVA ↔ amps — worked example

Example inputs

  • Value: 500
  • Conversion: kVA → amps
  • Phase: 3Φ (three phase)
  • System voltage: 415 V — industrial
  • Power factor: 0.8

Example result

  • Line current: 696 A
  • Apparent power: 500 kVA
  • Active power: 400 kW

Assumptions & limits

  • Phase system: 3Φ balanced at 415 V
  • Multiplier used: √3 × V
  • Assumes a balanced, linear load profile
  • Preliminary planning only. Final selection needs project-specific engineering verification.

METHODS & WORKED EXAMPLES

Understand the estimate.

These examples use the default inputs. Change the live controls above for your own scenario; the reference examples below remain fixed.

Three-phase voltage unbalance — worked example

Example inputs

  • L1–L2 voltage: 415 V
  • L2–L3 voltage: 410 V
  • L3–L1 voltage: 420 V

Example result

  • Average line voltage: 415 V
  • Maximum deviation: 5 V
  • Voltage unbalance: 1.2 %

Assumptions & limits

  • Unbalance (%) = 100 × maximum absolute deviation from the average / average.
  • Use line-to-line readings taken under the same operating condition.
  • No pass/fail limit or trip delay is selected. Check equipment limits and investigate supply conditions with a qualified engineer.
  • This method uses voltage magnitudes only; it is not an IEC negative-sequence calculation.
  • Preliminary planning only. Final selection needs project-specific engineering verification.

Method reference: US DOE — Energy Management for Motor-Driven Systems